Inconsistency between definitions of power and work in continuum mechanics?

[Klaus3]

The definitions of power and work in continuum mechanics are:

$$ W = \int T \cdot u dA + \int b \cdot u dV (1) $$


$$ P = \int T \cdot v dA + \int b \cdot v dV (2) $$

$u$ is the displacement vector, $v$ is the velocity, $T$ is the cauchy stress tensor, $b$ is the body force $A$ is area and $V$, volume. $W$ is work and $P$ is power

I tried to show that $P = \frac{dW}{dt}$, but failed

At first i thought integrating Eq (1) but i was intimidated by the amount of algebra necessary to put the derivative inside the integral, then i tried integrating Eq (2)

$$ \int Pdt = \int \int T \cdot v dAdt + \int \int b \cdot v dV dt(3) $$

Commuting the Space and time integrals and applying the definition of velocity

$$ \int Pdt = \int \int T \cdot \frac{du}{dt} dtdA + \int \int b \cdot \frac{du}{dt} dtdV (4) $$

$$ \int Pdt = \int \int T \cdot dudA + \int \int b \cdot dudV (5) $$

And here is where i'm stuck. Equation (5) doesn't seem to match with equation (1) and i don't know if there is any mistake in the derivation. Instead of $du$ it should have been $u$ in Eq (5), but a priori i don't know how to make it appear. Any clues? Thanks

 

[anuttarasammyak]

v=\frac{du}{dt}
Replacing u in W to v, we get P.

 

[Klaus3]

v=\frac{du}{dt}
Replacing u in W to v, we get P.

How? $v = \frac{du}{dt}$ but

$$ u = \int v dt $$

Then

$$ W = \int T \cdot \left( \int v dt\right) dA + \int b \cdot \left( \int v dt\right) dV $$


$$ \frac{dW}{dt} = \frac{d}{dt} \int T \cdot \left( \int v dt\right) dA + \frac{d}{dt} \int b \cdot \left( \int v dt\right) dV $$

Now what? I don't think i can simply "cancel" the outer time derivative with the inner time integral, can i?

 

[anuttarasammyak]

Use the law of differential of product
(ab)'=a'b+ab'
and, if I understand the settings correctly,
\frac{dT}{dt}=\frac{db}{dt}=0

 

[Klaus3]

Use the law of differential of product
(ab)'=a'b+ab'
and, if I understand the settings correctly,
\frac{dT}{dt}=\frac{db}{dt}=0

$\frac{dT}{dt}$ and $\frac{db}{dt}$ are generally not zero. For example, imagine $T$ is a pressure, pressure is not necessarily constant on a general process.

Also, the product rule isn't straightforward for volume and area integrals, because generally they also vary with time (which is what ends up in the reynolds transport theorem)

 

[anuttarasammyak]

Does your textbook, which describes the definition of W and P, postulate stationary T and b ?

 

[Klaus3]

It does not, they just define it that way.

 

[anuttarasammyak]

The latter refers the three statistical equations so it seems not a general relation.

 

[Klaus3]

I just couldn't find an explicit mention of work in the textbook (its always power), but its still the same definition found in other places, such as:

https://en.wikipedia.org/wiki/Virtu..._of_virtual_work_and_the_equilibrium_equation

We start by looking at the total work done by surface traction on the body going through the specified deformation:

i know the text talks about virtual work, but this excerpt specifically deals with regular work.

 

[anuttarasammyak]

The definitions of power and work in continuum mechanics are:

Who said these definitions in what conditions ? You may be haunted by a wrong idea.